Show that $$\lim_{t\to 0}\frac{1}{t}\int_2^\infty\frac{1-\cos(tx)}{x^2\log x}dx=0$$
This is arising as an intermediate step in a problem.
My attempts
I thought bounding this by another function which goes to zero with $t$ will help. But taking the trivial bound $|1-\cos x|\leq 2$ doesn't help. Taking another bound $|{1-\cos x}|\leq\frac{x^2}{2}$ also doesn't help, since $\int_2^\infty\frac{1}{\log x}dx$ is divergent.
Please help.
Note that we may assume $ t > 0$. Using the inequality that $\cos(x) \geq 1 - x^2/2$, we have
\begin{align*} 0 \leq \frac{1}{t} \int_{2}^{\infty} \frac{1-\cos(tx)}{x^2\log x} \, \mathrm{d}x &\leq \frac{1}{t} \int_{2}^{1/t} \frac{(tx)^2/2}{x^2\log x} \, \mathrm{d}x + \frac{1}{t} \int_{1/t}^{\infty} \frac{2}{x^2\log x} \, \mathrm{d}x \\ &= t \int_{2}^{1/t} \frac{\mathrm{d}x}{\log x} + \frac{1}{t} \int_{1/t}^{\infty} \frac{2}{x^2\log x} \, \mathrm{d}x. \tag{1} \end{align*}
For the first term, we can apply L'Hospital's theorem to get
$$ t \int_{2}^{1/t} \frac{\mathrm{d}x}{\log x} = \frac{\int_{2}^{1/t} \frac{\mathrm{d}x}{\log x}}{1/t} \sim \frac{-1/t^2\log(1/t)}{-1/t^2} = \frac{1}{\log(1/t)}. $$
Likewise, by the L'Hospital's theorem,
$$ \frac{\int_{1/t}^{\infty} \frac{2}{x^2\log x} \, \mathrm{d}x}{t} \sim \frac{1/(t^2(1/t)^2\log(1/t))}{1} = \frac{1}{\log (1/t)}. $$
Therefore the bound $\text{(1)}$ is $\mathcal{O}(1/\log(1/t))$ and converges to $0$ as $t \to 0^+$.